compound interest mcq pdf Model Questions & Answers, Practice Test for ibps rrb so officer sclae 2 3 single exam 2024

Answer: (c)

Principal + S.I. for 3$1/2$ years

= Rs. 7360.50 ....... (i)

Principal + S.I. for 2 years

= Rs. 6780 ...... (ii)

On subtracting equation (ii) from (i),

S.I. for 1$1/2$ years = Rs. 580.50

∴ S.I. for 2 years

= Rs. $({580.50 × 2 × 2}/3)$ = Rs. 774

∴ Principal = Rs. (6780 – 774 )

= Rs. 6006

And, rate of interest

= ${774 × 100}/{6006 × 2}$ = 6.4% per annum.

Answer: (b)

Answer: (b)

Amt = 10,000$(1 + 10/100)(1 + 12/100)$ = 12320

Answer: (c)

P = $A/(1 + r/100)^t$

Here, A = RS. 9261

r = 5% per annum

t = 3 years

P = $9261/(1 + 5/100)^3 = 9261/{9261/8000}$

P = 8000.

Answer: (b)

Let the sum borrowed at 6% be Rs. x = $P_1$

Then the sum borrowed at 10%

= Rs. (16000 – x ) = $P_2$

Time is one year in both cases

$R_1$ = 6%

$R_2$ = 10%

I = $I_1 + I_2$

I = ${P_1 R_1 T}/100 + {P_2 R_2 T}/100$

I = $T/100 (P_1 R_1 + P_2 R_2)$

or $P_1 R_1 + P_2 R_2 = {100 I}/T$

On substitution we get,

(x × 6) + (16000 – x)10

= ${100 × 1120}/1$

⇒ 160000 – 4x =112000

⇒ 4x = 48000

⇒ x = Rs. 12000

and 16000 – x = Rs. 4000.